Problem 39

Question

Use the Laplace transform to solve the given initial-value problem. $$ 2 y^{\prime \prime \prime}+3 y^{\prime \prime}-3 y^{\prime}-2 y=e^{-t}, \quad y(0)=0, y^{\prime}(0)=0, y^{\prime \prime}(0)=1 $$

Step-by-Step Solution

Verified

Answer

Solve for $ Y(s) $ using Laplace transforms and return to $ y(t) $.

1Step 1: Take the Laplace Transform of Both Sides

Apply the Laplace transform to each term of the differential equation $ 2y^{\prime \prime \prime} + 3y^{\prime \prime} - 3y^{\prime} - 2y = e^{-t} $. Use the properties of the Laplace transform: $\mathcal{L}\{ y^{(n)} \} = s^n Y(s) - s^{n-1} y(0) - \cdots - y^{(n-1)}(0)$.

2Step 2: Substitute Initial Conditions

Substitute the initial conditions $ y(0)=0 $, $ y^{\prime}(0)=0 $, $ y^{\prime \prime}(0)=1 $ into the transformed equation. This gives us an equation in terms of $ Y(s) $.

3Step 3: Simplify and Solve for $ Y(s) $

Combine all terms and solve for $ Y(s) $ by factoring and isolating $ Y(s) $ on one side of the equation after transforming the given terms.

4Step 4: Use Partial Fraction Decomposition

If necessary, use partial fraction decomposition to separate $ Y(s) $ into simpler terms that can be inverse transformed back into the time domain.

5Step 5: Apply Inverse Laplace Transform

Apply the inverse Laplace transform to the expression found for $ Y(s) $ to find $ y(t) $ in the time domain. Use known inverse transforms for common Laplace terms.

Key Concepts

Initial Value ProblemDifferential EquationsInverse Laplace TransformPartial Fraction Decomposition

Initial Value Problem

An initial value problem (IVP) involves a differential equation coupled with a specific set of conditions, known as initial conditions. These conditions specify the values of the solution and its derivatives at a particular point in time, often at the beginning, where time equals zero.
For example, in this problem:

The equation is: $2y''' + 3y'' - 3y' - 2y = e^{-t}$
Initial conditions are given as: $ y(0)=0, y'(0)=0, y''(0)=1 $

These initial conditions help tailor the general solution of the differential equation to the specific needs of the problem. By providing initial conditions, we ensure that the problem has a unique solution that fits the context in which it is applied.

Differential Equations

Differential equations involve relationships between a function and its derivatives. They are fundamental in expressing physical phenomena, such as motion, heat, or waves, mathematically. In this context, the differential equation $2y''' + 3y'' - 3y' - 2y = e^{-t}$ encompasses both sides:

The left side features derivatives of function $y(t)$ up to the third order.
The right side features an exponential term $e^{-t}$, representing an external forcing function.

Solving this equation means finding a function $y(t)$ that satisfies both its formal structure and its initial conditions. Differential equations can be solved using various techniques, and the Laplace transform is a particularly useful method for linear differential equations with constant coefficients.

Inverse Laplace Transform

The inverse Laplace transform is a key concept needed to revert back to the time domain from the complex frequency domain. When solving differential equations using Laplace transforms, we typically transform the original equation into an algebraic equation in terms of $Y(s)$. After finding $Y(s)$, it must be transformed back to ascertain the time-dependent function $y(t)$.
The steps are:

Transform $Y(s)$ using known inverse transforms.
Use any necessary tables or properties of the inverse Laplace transform.

This step allows us to interpret the solution in terms of time, making it physically meaningful. Without performing the inverse transform, we cannot obtain the actual solution that matches the initial conditions in the time domain.

Partial Fraction Decomposition

Partial fraction decomposition is a technique used to simplify complex rational expressions into sums of simpler fractions. This is especially useful in solving differential equations using the Laplace transform, as it allows easier application of inverse transforms.
Once $Y(s)$ is determined, it may take the form of a complicated rational expression:

Break the expression into simpler components.
Make each component amenable to quick inverse transformation.

Using partial fraction decomposition typically involves expressing a given fraction as a sum of terms, where each term has a polynomial of a lower degree in both numerator and denominator. This simplification is crucial as it significantly eases the process of applying the inverse Laplace transform by aligning each term with standard forms provided in transform tables.

Problem 39

Other exercises in this chapter

Problem 39

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Problem 39

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Problem 39

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Problem 39

Use the Laplace transform to solve each system. $$ \begin{aligned} &x^{\prime}+y=t \\ &4 x+y^{\prime}=0 \\ &x(0)=1, y(0)=2 \end{aligned} $$

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